In the reaction
$C_{2} O_{4}^{2 -} + M n O_{4}^{-} + H^{+} ⟶ M n^{2 +} + C O_{2} + H_{2} O$ the reductant is

$C_{2} O_{4}^{2 -}$

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$M n O_{4}^{-}$

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$M n^{2 +}$

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$H^{+}$

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Solution

The correct option is A
$C_{2} O_{4}^{2 -}$

$C_{2} O_{4}^{2 -} + M n O_{4}^{-} + H^{+} ⟶ M n^{2 +} + C O_{2} + H_{2} O$
In this reaction $C_{2} O_{4}^{2 -}$ act as a reducing agent.

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Similar questions

In the reaction
$C_{2} O_{4}^{2 -} + M n O_{4}^{-} + H^{+} ⟶ M n^{2 +} + C O_{2} + H_{2} O$ the reductant is

C_{2} O_{4}^{2 -} + M n O_{4}^{-} + H^{+} \to M n^{2 +} + C O_{2} + H_{2} O

In the reaction
$C_{2} O_{4}^{2 -} + M n O_{4}^{-} + H^{+} \to M n^{2 +} + C O_{2} + H_{2} O$ which is the reducing agent?

M n O_{4}^{2 -}, C_{2} O_{4}^{2 -}

H^{+}

M n O_{4}^{-} + C_{2} O_{4}^{2 -} + H^{+} \to M n^{2 +} + C O_{2} + H_{2} O

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